Hibernate: Parameter index out of range (8 > number of parameters, which is 7)

Question

Not a duplicate of this question Parameter index out of range (8 > number of parameters, which is 7)

My SaltTranDef entity class is

@Id
@Column(name="salt_id")
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer saltId;

@Column(name="tran_type")
private String transactionType;

@Column(name="user_id")
private String userId;

@Column(name="parent_system")
private String parentSystem;

@Column(name="parent_sys_ref_id")
private String parentSystemReference;

@Column(name="status")
private int status;

@OneToMany(mappedBy = "saltTranDef")
@Cascade({ org.hibernate.annotations.CascadeType.ALL,
         org.hibernate.annotations.CascadeType.DELETE_ORPHAN })
private Set<SaltTranUser> saltTranUsers;

And the SaltTranUser entity class is

@Id
@Column(name="salt_id")
private Integer saltId;

@Id
@Column(name="salt_property")
private String saltProp;

@Column(name="salt_value")
private String saltValue;

@ManyToOne
@JoinColumn(name="salt_id")
private SaltTranDef saltTranDef;

Both the above entity classes extend a mappedSuperclass

@Column(name="cre_suid")
private String creatorId;

@Column(name="mod_suid")
private String modifiedId;

@Column(name="cre_date")
private Timestamp creationDate;

@Column(name="mod_date")
private Timestamp modificationDate;

When inserting from a JUnit:

@Test
public void testInsert(){

    SaltTranDef std = new SaltTranDef();
    SaltTranUser stu1 = new SaltTranUser();
    SaltTranUser stu2 = new SaltTranUser();
    SaltTranUser stu3 = new SaltTranUser();
    Set<SaltTranUser> set1 = new HashSet<SaltTranUser>();

    Transaction tx = session.beginTransaction();

    std.setParentSystem("A");
    std.setParentSystemReference("123");
    std.setStatus(10);
    std.setTransactionType("A");
    std.setUserId("1234");
    std.setCreationDate(new Timestamp(new Date().getTime()));
    std.setCreatorId("1234");

    session.persist(std);
//  session.flush();

    stu1.setCreationDate(new Timestamp(new Date().getTime()));
    stu1.setCreatorId("1234");
    stu1.setSaltProp("Fname");
    stu1.setSaltValue("Swateek");
    stu1.setSaltId(std.getSaltId());

    stu2.setCreationDate(new Timestamp(new Date().getTime()));
    stu2.setCreatorId("1234");
    stu2.setSaltProp("Lname");
    stu2.setSaltValue("Jena");
    stu2.setSaltId(std.getSaltId());

    stu3.setCreationDate(new Timestamp(new Date().getTime()));
    stu3.setCreatorId("1234");
    stu3.setSaltProp("Phone");
    stu3.setSaltValue("9900056668");
    stu3.setSaltId(std.getSaltId());

    set1.add(stu1);
    set1.add(stu2);
    set1.add(stu3);

    std.setSaltTranUsers(set1);

    session.save(std);
    tx.commit();
}

I get an error saying:

SEVERE: Parameter index out of range (8 > number of parameters, which is 7). Mar 25, 2015 8:06:35 AM org.hibernate.event.def.AbstractFlushingEventListener performExecutions SEVERE: Could not synchronize database state with session org.hibernate.exception.GenericJDBCException: could not insert: [com.salt.entity.SaltTranUser] at org.hibernate.exception.SQLStateConverter.handledNonSpecificException(SQLStateConverter.java:103) Caused by: java.sql.SQLException: Parameter index out of range (8 > number of parameters, which is 7). at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1055)

Putting @Transient before @ManyToOne worked fine for me. — rajya vardhan
– rajya vardhan, Commented Dec 7, 2017 at 11:55

Radim Köhler · Accepted Answer · 2015-03-25 05:30:41Z

7

This kind of problem is almost always related to double column mapping. And indeed. We can see, that this mapping uses one column twice "salt_id":

the SaltTranUser entity class:

@Id
@Column(name="salt_id")
private Integer saltId;
...

@ManyToOne
@JoinColumn(name="salt_id")
private SaltTranDef saltTranDef;

And that is wrong. Hibernate is at the end inserting into one column twice, i.e. more arguments then columns in INSERT, UPDATE

Solution here would be mostlikely very simple - because the @ManyToOne seems to be wrong. I would expect some special column for reference like: SaltTranDef_id

answered Mar 25, 2015 at 5:30

Radim Köhler

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5 Comments

swateek Over a year ago

This approach I can't work out due to some business reasons. So, in the second table the Identity column of first table and another column would be a composite primary key. The identity column in the second table would be a foreign key dependent on the first table.

Radim Köhler Over a year ago

Mark it as read-only. But in general, my answer is THE answer. Two mappings of one column cannot be used as writeable... Hope it helps

swateek Over a year ago

Now..with some changes in the entity class.. (I have done that read only too) this error isn't coming. But something else is coming up, I am confused if I should edit this question or start a different thread? Help please.

Radim Köhler Over a year ago

To get more attention,ask new question. This is down in the stack.

swateek Over a year ago

Here's the new question that I have posted, there are a few changes in the entity classes too. If you could please help. stackoverflow.com/questions/29285225/…

message · Accepted Answer · 2016-09-07 13:49:22Z

4

In my case issue was with user_type field. It was defined as discriminator and as public property. It wasn't marked with @Column but somehow Hibernate anyway failed on this

@Entity
@Table(
    name = "abstract_users",
    indexes = {@Index(name = "idx__abstract_users__is_enabled", columnList = "is_enabled")}
)
@Access(AccessType.FIELD)
@Inheritance(strategy = InheritanceType.JOINED)
// @TODO: rename user_type to discriminator
@DiscriminatorColumn(name = "user_type", discriminatorType = DiscriminatorType.STRING, length = 10)
abstract public class AbstractUser extends CreateUpdateTimestampableBase
{

    // @TODO: rename user_type to discriminator
    public String user_type;

    @Min(0)
    @NotNull
    @Column(precision = 19, scale = 2)
    protected BigDecimal balance;

    //...
}

answered Sep 7, 2016 at 13:49

message

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2 Comments

CRitchie Over a year ago

Hibernate picks up any property that is not marked transient or annotated with @Transient. That is why user_type detected as a database column.

Albert Waninge Over a year ago

Can we say that Hibernate does not allow you to use the discriminator column as a property of the class?

Pankaj Shinde · Accepted Answer · 2017-10-31 06:50:42Z

4

I faced same issue when I was using the one-directional mapping (parent class containing child class, but child class doesn't keep the reference of parent class). The mapping looked like

@OneToMany(cascade = CascadeType.ALL, fetch=FetchType.EAGER, orphanRemoval = true)
@JoinColumn(name="jobcard_id", nullable=false)
private List<JobServiceMapping> services;

I got the error Parameter index out of range. Then I changed the annotations a bit and now it is working for me.

@OneToMany(mappedBy="jobcardId", cascade = CascadeType.ALL, fetch=FetchType.EAGER, orphanRemoval = true)
private List<JobServiceMapping> services;

answered Oct 31, 2017 at 6:50

Pankaj Shinde

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Comments

Lucas Besnard · Accepted Answer · 2020-05-09 20:38:08Z

0

In your entity, replace

@ManyToOne
@JoinColumn(name="salt_id")
private SaltTranDef saltTranDef;

by

@ManyToOne
@JoinColumn(name="salt_id", updatable = false, insertable = false)
private SaltTranDef saltTranDef;

it worked for me

answered May 9, 2020 at 20:38

Lucas Besnard

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